算法笔记上机训练实战指南第10章提高篇(4)--图算法专题学习笔记

10.3 图的遍历

PAT A1013 Battle Over Cities (25分)

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.

For example, if we have 3 cities and 2 highways connecting

Input Specification:

Each input file contains one test case. Each case starts with a line containing 3 numbers

Output Specification:

For each of the

Sample Input:

Sample Output:

1
0
0

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int n,m,k;
const int N = 1111;
vector<int> G[N];
bool vis[N];
int currentPoint;
void dfs(int v){
    if(v == currentPoint)
        return;
    vis[v] = true;
    for(int i=0;i<G[v].size();i++){
        if(vis[G[v][i]] == false)
            dfs(G[v][i]);
    }
}
int main(){
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        G[b].push_back(a);
    }
    for(int query = 0;query<k;query++){
        scanf("%d",&currentPoint);
        memset(vis,false,sizeof(vis));
        int block = 0;
        for(int i=1;i<=n;i++){
            if(i != currentPoint && vis[i]==false){
                dfs(i);
                block++;
            }
        }    
        printf("%d\n",block-1);
    }
    return 0;
}

PAT A1021 Deepest Root (25分)

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print Error: K components where K is the number of connected components in the graph.

Sample Input 1:

Sample Output 1:

3
4
5

Sample Input 2:

Sample Output 2:

Error: 2 components

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int N = 10010;
vector<int> G[N];
bool isRoot[N];
int father[N];
int findFather(int x){
    int a = x;
    while(x != father[x]){
        x = father[x];
    }
    while(a != father[a]){
        int z = a;
        a = father[a];
        father[z] = x;
    }
    return x;
}
void Union(int a,int b){
    int faA = findFather(a);
    int faB = findFather(b);
    if(faA != faB){
        father[faA] = faB;
    }
}
void init(int n){
    for(int i=1;i <= n;i++){
        father[i] = i;
    }
}
int calBlock(int n){
    int Block = 0;
    for(int i = 1; i <= n;i++){
        isRoot[findFather(i)] = true;
    }
    for(int i = 1; i <= n ;i++){
        Block += isRoot[i];
    }
    return Block;
}
int maxH = 0;
vector<int> temp,Ans;
void DFS(int u,int Height,int pre){
    if(Height > maxH){
        temp.clear();
        temp.push_back(u);
        maxH = Height;
    }else if(Height == maxH){
        temp.push_back(u);
    }
    for(int i=0;i<G[u].size();i++){
        if(G[u][i] == pre)
            continue;
        DFS(G[u][i],Height+1,u);
    }
}
int main(){
    int a,b,n;
    scanf("%d",&n);
    init(n);
    for(int i=1;i<n;i++){
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        G[b].push_back(a);
        Union(a,b);
    }
    int Block = calBlock(n);
    if(Block != 1){
        printf("Error: %d components\n",Block);
    }else{
        DFS(1,1,-1);
        Ans = temp;
        DFS(Ans[0],1,-1);
        for(int i = 0;i < temp.size();i++){
            Ans.push_back(temp[i]);
        }
        sort(Ans.begin(),Ans.end());
        printf("%d\n",Ans[0]);
        for(int i = 1; i < Ans.size(); i++){
            if(Ans[i] != Ans[i-1]){
                printf("%d\n",Ans[i]);
            }
        }
    }
}